Recent content by cse63146

I figured it out. Thank you all for your help.

That would mean I'm counting the pairs twice. So it would be: (ie 4 Choose 2)/2 = 3 ways of choosing?

So would a combination fit more than a permutation (ie 4 Choose 2)?

P(2 pairs) = 4!(0.1)2(0.9). That makes sense. Thank you. Just one more question. Would the probability of getting 1 pair be: 4C2P(second card matches 1st)P(3rd card is different from 1st and 2nd)P(4th different from 3rd) P(pair) = 4C2(0.1)(0.9)(0.8)?

It's not really a deck per say. You are dealt a card from a stack of 10 cards (0-9), record the number, return the card back to the deck,shuffle, and draw again. Do this until you recorded 4 numbers. The probability of getting any number is 0.1.

Homework Statement You are dealt 4 cards (each card is numbered from 0 - 9). Each card is independed of the other card. What is the probability of getting 2 pairs? Homework Equations The Attempt at a Solution P(2 pair) = *4C2P(1st = 2nd)P(3rd ≠ 2nd = 1st) P(4th = 3rd ≠ 2nd =...

I was thinking of doing something like that, but the question says that I have to use the fact that the interior angle of a triagngle sum up to 180. Maybe it would help if I move the a*b* line somewhere else, hopefully it'll work.

Maybe this would help. The first part is to use Parallel Transport to prove that the sum of the interior angles sum up to 180. Now I have to do the opposite.

The problem is: Using the fact that the interior angles of a triangle sum up to 180, prove parallel transport (or PT!).

I'm trying to use the fact that a + b + c = 180, and somehow prove that ab and a*b* are parallel. If I can show that the two angles are congruent to another, then the lines would be parallel.

Homework Statement [PLAIN]http://img831.imageshack.us/img831/2589/88722757.jpg Using only the fact that the sum of the interior angles of a triangle is 180, how would I show that a = a* and b = b*? Homework Equations The Attempt at a Solution In my diagram, there are 2...

I did it somewhat different. Let T = Never reach State 2 U_i = P(T|X_0 = i) for i = 0,1,2,3 U_1 = 1 and U_2 = 0. This would just leave U_1 and U_3 (two unknowns) and also 2 equations. I would solve for both of them, and U_1 would give me the desired probability. Would that approach work...

Yes, it is. S = {0,1,2,3}, and the matrix is \begin{bmatrix}1 & 0 & 0 & 0 \\ 0.1 & 0.2 & 0.5& 0.2 \\ 0.1 & 0.2 & 0.6 & 0.1 \\ 0.2 & 0.2 & 0.3 & 0.3\end{bmatrix} where the columns are (0,1,2,3) and the rows are (0,1,2,3)'. Do you know how to find the probability that it never reaches state 2?

Homework Statement Given a Probability transition matrix, starting in X0= 1, determine the probability that the process never reaches state 2. Homework Equations The Attempt at a Solution State 2 is not an observing state, so I'm not sure how to find this probability. Any help...

Homework Statement A card is picked at random from N cards labeled 1,2,...,N and the number that appears is X. A second card is picked at random from cards numbered 1,2,..., X and its number is Y. Find the conditional distribution of X given Y = y. Homework Equations P(X = x | Y = y) =...